The slope of the tangent to a curve at any point (x, y) is given by (y^2+1)^2 1 - 2xy(y^2+1) . If the curve passes through the origin, then the area of the region bounded by the curve, the y-axis, and the line y = 1 is

Step-by-step solution

Given the slope of the tangent is dy dx = (y^2+1)^2 1 - 2xy(y^2+1) Taking the reciprocal, we get: dx dy = 1 - 2xy(y^2+1) (y^2+1)^2 dx dy = 1 (y^2+1)^2 - 2y y^2+1 x dx dy + ( 2y y^2+1 )x = 1 (y^2+1)^2 This is a linear differential equation of the form dx dy + P(y)x = Q(y) . Integrating Factor (I.F.) = e^ 2y y^2+1 dy = e^ (y^2+1) = y^2+1 The general solution is: x (y^2+1) = 1 (y^2+1)^2 (y^2+1) dy + C x(y^2+1) = 1 y^2+1 dy + C x(y^2+1) = ⁻¹y + C Since the curve passes through the origin (0,0) , we have 0 = ⁻¹(0) + C C